7 lines
547 B
Markdown
7 lines
547 B
Markdown
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To be a permutation all six numbers have to have the same digit count. So `digitcount(x) == digitcount(x*6)`.
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This for each number of digits only given for the numbers from `10^n` to `10/6 * 10^n`.
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We perform the permutation check with an modified version of the algorithm used in problem-49 *(product of all digits plus two)*.
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But we generalise the code to work with every number of digits.
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And because we greatly limited the amount of numbers to search and the permutation test is pretty fast this is all we need to do (except run the code).
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